On Mon, 07 Jun 2004 22:22:49 -0500, Cecil Moore wrote:

Walter Maxwell wrote:
The equation he misunderstands is Eq 4.23 on Page 100 in Johnson, which is
explained and derived on Pages 98 and 99, to derive the voltage E of the
standing wave for any position along a mismatched transmission line. Steve's
misunderstanding is that he believes the equation expresses the value of the
forward voltage on the line. Consequently, his Eq 6 in Part 1 says that 'Vfwd =
the terms on the right-hand side of the equation copied from Johnson', while the
correct version is 'E = the terms on the righ-hand side'.

Johnson's equation contains all possible components - both forward and
reflected, so Johnson's equation predicts the total net standing-wave
voltage. (Note that Johnson's equation contains both positive and negative
exponents.)

Dr. Best's equation contains only the forward components of Johnson's
equation. Therefore, Dr. Best's equation predicts *only* the forward-
traveling voltage. (Note that Dr. Best's equation contains only negative
exponents and represents the E+ half of Johnson's equation while omitting
the E- half of Johnson's equation.)

In other words, the two equations do *not* predict the same quantity
and they are *not* supposed to be the same equations. As I said before,
Dr. Best made a lot of conceptual errors but his equations seem to be
valid. If you break Johnson's equation into two parts representing E+
and E-, the E+ part will be Dr. Best's Eq 6.

Please consider the following matched system with a 1/2 second long lossless
transmission line. The physical rho is 0.5. Assume VF1 is a constant 70.7V.

100W XMTR---50 ohm line---x---1/2 second long lossless 150 ohm line---50 ohm load
VF1=70.7V-- VF2--
--VR1 --VR2

V1 is equal to 70.7(1.5) = 106.06V and is a constant value
V2 starts out at zero and builds up to 35.35V
VF2=V1+V2 assuming they are in phase

Here is (conceptually simplified) how VF2 builds up to its steady-state value.
VR2(t+1) is 1/2 of VF2(t) and V2(t+1) is 1/4 of VF2(t)

Time in
Seconds V1 VR2 V2 VF2
0 106.06 0 0 106.06
1 106.06 53.03 26.5 132.56
2 106.06 66.26 33.13 139.19
3 106.06 69.58 34.79 140.85
4 106.06 70.4 35.2 141.26
5 106.06 70.63 35.32 141.38
6 106.06 70.69 35.34 141.39
7 106.06 70.7 35.35 141.4 very close to steady-state

My rounding is a little off but above is a close approximation to what happens.
Note that VF2 is equal to V1 + V2 and converges on the proper value of 141.4V.

Sorry to spoil your fun, Cecil, but 141.4 v is not the forward voltage, it's the
max voltage of the standing wave. This is the same mistake that Steve made. I
asked you to rethink what the forward voltage and you have come up wrong.

You are using circuit theory for superposition, as Steve did, when circuit
theory fails to apply in certain transmission line cases. This is one of those
cases. In circuit theory you can superpose the voltages from two sources and V1
+ V2 equals Vtotal. But the re-reflected voltage CANNOT be added to the source
voltage in the transmission line case to obtain Vfwd, because there is only ONE
source.

I'll say it again, Cecil, V1 + V2 = maxV of the standing wave--V1 + V2 does NOT
equal Vfwd because V2 is not a second source, it came fr om ONE source, the
transceiver, and therefore superposition of V1 and V2 does not apply to
establish Vfwd.

Therefore, it is true that maxV may be incident on the load if the relative
phase between the reflected and forward waves permits, The only time the forward
wave is incident on the load is when the load = Zo.

Please, Cecil, go back to the drawing board and come up with the correct
Vfwd--it is not equal in any way to the right-hand side of Johnson's Eq 4.23;.
Remember, I said earlier that when rho = 0.5 and the circuit is matched, Vfwd =
Vsource x 1.1547. When you discover where the 1.1547 came from when rho = 0.5
you will have discovered the source of Vfwd.

Steady-state VF2 = 106.06 + 26.5 + 6.63 + 1.66 + 0.41 + ...
This is indeed of the form 106.06(1 + 1/4 + 1/16 + 1/64 + 1/256 + ...)
or VF2 = V1(1 + A + A^2 + A^3 + A^4 + ...)

Wrong, Cecil. Change VF2 to 'E', the max value of the standing wave, and the
values obtained from the sum of the terms in the geometric series will be
correct.

Walt

On Tue, 08 Jun 2004 15:54:50 GMT, Walter Maxwell wrote:

On Mon, 07 Jun 2004 22:22:49 -0500, Cecil Moore wrote:

Walter Maxwell wrote:
The equation he misunderstands is Eq 4.23 on Page 100 in Johnson, which is
explained and derived on Pages 98 and 99, to derive the voltage E of the
standing wave for any position along a mismatched transmission line. Steve's
misunderstanding is that he believes the equation expresses the value of the
forward voltage on the line. Consequently, his Eq 6 in Part 1 says that 'Vfwd =
the terms on the right-hand side of the equation copied from Johnson', while the
correct version is 'E = the terms on the righ-hand side'.

snip
Sorry to spoil your fun, Cecil, but 141.4 v is not the forward voltage, it's the
max voltage of the standing wave. This is the same mistake that Steve made. I
asked you to rethink what the forward voltage and you have come up wrong.

You are using circuit theory for superposition, as Steve did, when circuit
theory fails to apply in certain transmission line cases. This is one of those
cases. In circuit theory you can superpose the voltages from two sources and V1
+ V2 equals Vtotal. But the re-reflected voltage CANNOT be added to the source
voltage in the transmission line case to obtain Vfwd, because there is only ONE
source.

I'll say it again, Cecil, V1 + V2 = maxV of the standing wave--V1 + V2 does NOT
equal Vfwd because V2 is not a second source, it came fr om ONE source, the
transceiver, and therefore superposition of V1 and V2 does not apply to
establish Vfwd.

I'll make it a little easier for you, Cecil. I' ll give you two ways to
determine Vfwd.

Way 1. Vfwd = sqrt(Power fwd x Zo)

Way 2. Vfwd = Vsource x sqrt[1/(1- rho^2)]

Now plug rho = 0.5 into the expression for Way 2 and see what comes up.

Walt

Therefore, it is true that maxV may be incident on the load if the relative
phase between the reflected and forward waves permits, The only time the forward
wave is incident on the load is when the load = Zo.

Please, Cecil, go back to the drawing board and come up with the correct
Vfwd--it is not equal in any way to the right-hand side of Johnson's Eq 4.23;.
Remember, I said earlier that when rho = 0.5 and the circuit is matched, Vfwd =
Vsource x 1.1547. When you discover where the 1.1547 came from when rho = 0.5
you will have discovered the source of Vfwd.

Steady-state VF2 = 106.06 + 26.5 + 6.63 + 1.66 + 0.41 + ...
This is indeed of the form 106.06(1 + 1/4 + 1/16 + 1/64 + 1/256 + ...)
or VF2 = V1(1 + A + A^2 + A^3 + A^4 + ...)

Wrong, Cecil. Change VF2 to 'E', the max value of the standing wave, and the
values obtained from the sum of the terms in the geometric series will be
correct.

Walt

I'll make it a little easier for you, Cecil. I' ll give you two ways to
determine Vfwd.

Way 1. Vfwd = sqrt(Power fwd x Zo)

Way 2. Vfwd = Vsource x sqrt[1/(1- rho^2)]

Now plug rho = 0.5 into the expression for Way 2 and see what comes up.

Walt

Cecil, let's take it one step further and include current, forward current, that
is, Ifwd.

Ifwd = Isouce x sqrt[1/(1 - rho^2)]

Now let's see what happens with a souce voltage of 70.71 v at 100 w on a 50-ohm
line. The source current is 1.414 a.

Now find Ifwd x Vfwd and see what comes up.

Walt

Walter Maxwell wrote:
Cecil, let's take it one step further and include current, forward current, that
is, Ifwd.

Ifwd = Isouce x sqrt[1/(1 - rho^2)]

This equation for voltage was wrong. Perhaps, we should resolve that
problem first.

Forward current can be calculated from the square root of (forward
power divided by Z0). So IF2 = SQRT(133.33W/150ohms) = 0.943 amps

Funny thing is that 141.4v*0.943a = 133.33 watts, the known forward
power on the 150 ohm line.
--
73, Cecil http://www.qsl.net/w5dxp



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Walter Maxwell wrote:
I'll make it a little easier for you, Cecil. I' ll give you two ways to
determine Vfwd.

OK, here's the example:

100W XMTR---50 ohm line---x----1/2WL 150 ohm line---50 ohm load
PF1=100W-- PF2=133.33W--
--PR1=0W --PR2=33.33W

Way 1. Vfwd = sqrt(Power fwd x Zo)

So Vfwd = sqrt(133.33W x 150ohms) = 141.4 volts (that's what I said)

Way 2. Vfwd = Vsource x sqrt[1/(1- rho^2)]
Now plug rho = 0.5 into the expression for Way 2 and see what comes up.

OK, Vfwd = 70.7V x sqrt[1/0.75] = 81.6 volts (something wrong there)
We know Vfwd has to be 141.4V as calculated above using Ohm's law.

Walt, your two ways don't yield the same value. Perhaps that is the
source of your confusion about what Dr. Best said. Your Way 2 appears
to be a calculation of the voltage delivered to a mismatched load.
--
73, Cecil http://www.qsl.net/w5dxp




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On Tue, 08 Jun 2004 13:29:02 -0500, Cecil Moore wrote:

Walter Maxwell wrote:
I'll make it a little easier for you, Cecil. I' ll give you two ways to
determine Vfwd.

OK, here's the example:

100W XMTR---50 ohm line---x----1/2WL 150 ohm line---50 ohm load
PF1=100W-- PF2=133.33W--
--PR1=0W --PR2=33.33W

Way 1. Vfwd = sqrt(Power fwd x Zo)

So Vfwd = sqrt(133.33W x 150ohms) = 141.4 volts (that's what I said)

Way 2. Vfwd = Vsource x sqrt[1/(1- rho^2)]
Now plug rho = 0.5 into the expression for Way 2 and see what comes up.

OK, Vfwd = 70.7V x sqrt[1/0.75] = 81.6 volts (something wrong there)
We know Vfwd has to be 141.4V as calculated above using Ohm's law.

Walt, your two ways don't yield the same value. Perhaps that is the
source of your confusion about what Dr. Best said. Your Way 2 appears
to be a calculation of the voltage delivered to a mismatched load.

Cecil, you've hit the problem on the nose. Until a few posts past I didn't
realize you were using a 150-ohm line. I've been using 50 ohms for the line Zo
terminated with 150 ohms, and you'll find my numbers are correct for that Zo.
Vfwd = 81.6 v is correct for 100 w on a matched 50 ohm line with 3:1 mismatched
load. You'll also get 81.6 using my Way 1.

However, on a 50-ohm line with a 3:1 mismatch and Vfwd = 81.6v (actually it's
81.65 v), the reflected wave is 40.83 v. When the system is matched the phase of
the re-reflected wave is 0° and is also 40.83 v. Therefore, V1 = 81.65 v and V2
= 40.83,v, which means V1 + V2 cannot equal Vfwd. Yet, in the paragraph
preceding Steve's Eq 13 he says that when the phase is 0° Eq. 13 derives
PFtotal. It does not.

Using his Eq 7 V1 yields P1 = 133.33 w, and using his Eq 8 V2 P2 = = 33.33 w.
Using these values of P1 and P2 in his Eq. 12 yields PFtotal = 238 watts, which
is also what we get when plugging V1 and V2 into his Eq 9. We know 238 watts is
incorrect, because it need to be 133.33 w. It makes no sense to plug P1 from Eq
7 into Eq 12 to find PFtotal, because P1 is already PF total.

This clearly proves that Eq 12 is invalid, because Eq 7 and 8 are valid.

I am studying your case with the 150-ohm line, but I haven't yet corralled it.

Walt




re-reflected voltage is also 40.83 v.

Walter Maxwell wrote:
Cecil, you've hit the problem on the nose. Until a few posts past I didn't
realize you were using a 150-ohm line.

I've only explained and presented it about ten times now.

The impedances are irrelevant, Walt. The equations have to
work no matter what the impedances.

I have presented ASCII schematics of what I was discussing.
Would you be so kind as to do likewise? I am getting sick
and tired of this run-around.
--
73, Cecil http://www.qsl.net/w5dxp



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Walter, W2DU wrote:
"I`ll give you two ways to determine Vfwd."

Walter makes it too easy by using the power relationships, but that`s
the way it is. The forward (incident) wave is opposed by Ro in a
practical line. The reward (reflected) wave is also opposed by Ro (the
surge impedance) of the line.

Power from the transmitter is nearly the same as that delivered to the
load as loss is small and no significant room exists in the line to
store RF.

Power to the load is the difference between forward power and reflected
power.

The voltage at any point on a mismatched line is the sum of the forward
and reflected waves at that point but is merely a manifestatoion of SWR
and has little practical value. At a current or a voltage loop
(maximum), the forward and reflected amps or volts are in-phase. So, if
a line is opened at a loop point, the impedance looking toward the load
is a pure resistance regardless of the nature of the load (see page 37
of "Transmission Lines, Antennas, and Wave Guides" by King, Mimno, and
Wing).

Best regards, Richard Harrison, KB5WZI

Richard Harrison wrote:
The voltage at any point on a mismatched line is the sum of the forward
and reflected waves at that point but is merely a manifestatoion of SWR
and has little practical value.

All true and completely irrelevant to Dr. Best's article since he was
*never* talking about the net voltage. He is talking about the forward
voltage. Any notion that Dr. Best was talking about net voltage is a
mistaken notion. (I can't believe I'm defending Dr. Best but he said
what he said.)
--
73, Cecil http://www.qsl.net/w5dxp



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On Tue, 8 Jun 2004 13:56:06 -0500, (Richard Harrison)
wrote:

Walter, W2DU wrote:
"I`ll give you two ways to determine Vfwd."

Walter makes it too easy by using the power relationships, but that`s
the way it is. The forward (incident) wave is opposed by Ro in a
practical line. The reward (reflected) wave is also opposed by Ro (the
surge impedance) of the line.

Power from the transmitter is nearly the same as that delivered to the
load as loss is small and no significant room exists in the line to
store RF.

Power to the load is the difference between forward power and reflected
power.

The voltage at any point on a mismatched line is the sum of the forward
and reflected waves at that point but is merely a manifestatoion of SWR
and has little practical value. At a current or a voltage loop
(maximum), the forward and reflected amps or volts are in-phase. So, if
a line is opened at a loop point, the impedance looking toward the load
is a pure resistance regardless of the nature of the load (see page 37
of "Transmission Lines, Antennas, and Wave Guides" by King, Mimno, and
Wing).

Best regards, Richard Harrison, KB5WZI

Thanks, Richard, what you say is true, but problem is in determining the method
to find the forward voltage when the line is mismatched.

Walt